The orbit of geostationary satellite is circu | vedclass

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Question

$\frac{\mathrm{GMm}}{(\mathrm{R}+\mathrm{h})^{2}}=\mathrm{m} \omega^{2}(\mathrm{R}+\mathrm{h})$

$\omega=\sqrt{\frac{\mathrm{GM}}{(\mathrm{R}+\mathr

Vedclass · Accepted Answer

$(ii), (iii)$ and $(iv)$

Vedclass · Answer

$\frac{\mathrm{GMm}}{(\mathrm{R}+\mathrm{h})^{2}}=\mathrm{m} \omega^{2}(\mathrm{R}+\mathrm{h})$

$\omega=\sqrt{\frac{\mathrm{GM}}{(\mathrm{R}+\mathrm{h})^{3}}}=\frac{2 \pi}{\mathrm{T}}$

$T$ depends upon mass of earth's, radius of orbit.

Option is $( 4)$

The orbit of geostationary satellite is circular, the time period of satellite depends on $(i)$ mass of the satellite $(ii)$ mass of the earth $(iii)$ radius of the orbit $(iv)$ height of the satellite from the surface of the earth

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